# Exercise 7-9

(difficult) The implementation of nrand in S7.4.4/135 will not work for arguments greater than RAND_MAX. Usually, this restriction is no problem, because RAND_MAX is often the largest possible integer anyway. Nevertheless, there are implementations under which RAND_MAX is much smaller than the largest possible integer. For example, it is not uncommon for RAND_MAX to be 32767 (2^15 -1) and the largest possible integer to be 2147483647 (2^31 -1). Reimplement nrand so that it works well for all values of n.

# Solution

This exercise is great – it gives us the chance to create and understand an algorithm that generate random number beyond RAND_MAX (the largest default C++ random number that can be “reached” by the rand() function) whilst ensuring even probability distribution of that random number (i.e. the number “10” having the same chance of getting picked as the number “234”, etc.).

Andrew and Barbara (the authors) provided the original code for this exercise on GitHub here back in year 2000. I took the liberty to read through it with the aim of understanding and making sense of the solution. After 3 evenings of sketching diagrams in my notepad (and getting stuck multiple times), I eventually understood the underlying principle / concepts, and managed to come up with an algorithm built on the authors’ original solution (with slight differences).

In this post I shall illustrate the derivation and implementation of the algorithm (which I can actually visualise). I use PowerPoint to draw out some diagrams to aid visualising the algorithms. I would like to express my sincerest thanks to Andrew and Barbara for this very wonderful exercise.

## What Will Our nrand() Do?

We are going to create a function nrand() that returns a random integer, given the asymmetric range [0, randomOutcomeLimit).

e.g. if randomOutcomeLimit is 4, then the randomOutcome may be either 0, 1, 2, 3 – each with an equal (25%) chance of getting picked.

## Terminologies

Before going any further, I believe it is a good idea to define some core variables / terminologies.

• randomDriver: this is essentially rand(). The C++ default (seeding) random number that span within the asymmetric range [0, randomDriverLimit). We use randomDriver to “drive” (or compute) the eventual randomOutcome.
• randomDriverLimit: this is RAND_MAX + 1. i.e. “one-above” the maximum possible randomDriver value.
• randomOutcome: this is the random number that gets returned. It spans within the symmetric range [0, randomOutcomeLimit). The algorithm ensures that every single randomOutcome has the same probability of getting randomly selected. The randomOutcome is “driven” (or determined) by the randomDriver.
• randomOutcomeLimit: “one-above” the maximum possible randomOutcome value.
• bucketSize: this is “the glue” between randomDriver and randomOutcome. Think of it as the ratio between the two. To enable us to translate a randomDriver into a randomOutcome (and vice versa).
• remainder: this is used in the computation of randomOutcome. Only relevant for the case whhen randomOutcomeLimit > randomDriverLimit.

Note: in the “pseudo diagrams” that I have drawn up in PowerPoint, I shall use RD to represent randomDriver, and RO to represent randomOutcome.

## Algorithm

The algorithm for nrand() will cater for two scenarios:

• Case 1: when randomOutcomeLimit <= randomDriverLimit
• Case 2: when randomOutcomeLimit > randomDriverLimit

It may be easier if I display the complete nrand.cpp code here now, and walk through in more details afterwards.

### The nrand.cpp Code

#include <cmath>         // ceil
#include <cstdlib>       // rand, RAND_MAX, srand
#include <stdexcept>     // domain_error

using std::ceil;
using std::rand;
using std::domain_error;

// Asymmetric ranges:
//   randomOutcome: [0, randomOutcomeLimit)
//   randomDriver:  [0, randomDriverLimit)

int nrand(int randomOutcomeLimit) {

if (randomOutcomeLimit <= 0)
throw domain_error("Argument to nrand is out of range");

int randomOutcome;
int randomDriverLimit = RAND_MAX + 1;

if (randomOutcomeLimit <= randomDriverLimit) {
const int bucketSize = randomDriverLimit / randomOutcomeLimit;

do {
int bucket = rand() / bucketSize;
randomOutcome = bucket;
} while (randomOutcome >= randomOutcomeLimit);

} else {
const int bucketSize = ceil(randomOutcomeLimit / randomDriverLimit);

do {
int bucket = nrand(randomDriverLimit);
int remainder = nrand(bucketSize);
randomOutcome = (bucket * bucketSize) + remainder;
} while (randomOutcome >= randomOutcomeLimit);
}
return randomOutcome;
}


## Case 1: when randomOutcomeLimit <= randomDriverLimit


if (randomOutcomeLimit <= randomDriverLimit) {
const int bucketSize = randomDriverLimit / randomOutcomeLimit;

do {
int bucket = rand() / bucketSize;
randomOutcome = bucket;
} while (randomOutcome >= randomOutcomeLimit);

}



Since the values of randomOutcomeLimit and randomDriverLimit are pre-defined, we compute our bucketSize by taking the ratio between the two. We use the randomDriverlimit as the numerator (and randomDriverLimit as the denominator) to ensure this ratio is a non-fraction. Think of bucketSize as: “The number of randomDrivers per randomOutcome”. 10 randomDrivers per randomOutcome makes sense. 0.5 randomDrivers per randomOutcome does not.

Note that the equation automatically “round-down” the ratio to the integer (instead of “round-up”). This is to ensure every single randomOutcome to have the same probability of being picked.

To illustrate this, assume we have randomDriverLimit = 5 (i.e. RAND_MAX = 4. i.e. randomDriver may be 0, 1, 2, 3.), and randomOutcomeLimit = 2 (i.e. randomOutcome maybe 0 and 1).

### Compute bucketSize With “Round-down” (desirable)

In the case of computing bucketSize using “round-down”, the probability distribution of randomOutcome is even. This is what we want in a random number generator.

### Compute bucketSize With “Round-up” (Un-desirable)

In the case of computing bucketSize using “round-up”, the probability distribution of randomOutcome is non-even. This is what we DO NOT want in a random number generator.

### Further Visualisation

To visualise randomDriver, randomDriverLimit, randomOutcome, randomOutcomeLimit, bucket, and bucketSize all in one go, I have put together some PowerPoint diagrams here which I believe will speak for themselves without the need of much further (wordy) explanations! I assume a fixed randomDriverLimit of 21, and vary the randomOutcomeLimit for illustrations.

Visualise: randomDriverLimit of 21, randomOutcomeLimit = 2.

Visualise: randomDriverLimit of 21, randomOutcomeLimit = 3.

Visualise: randomDriverLimit of 21, randomOutcomeLimit = 4.

Visualise: randomDriverLimit of 21, randomOutcomeLimit = 5.

Visualise: randomDriverLimit of 21, randomOutcomeLimit = 6.

Visualise: randomDriverLimit of 21, randomOutcomeLimit = 7.

Visualise: randomDriverLimit of 21, randomOutcomeLimit = 11.

## Case 2: when randomOutcomeLimit > randomDriverLimit

  } else {
const int bucketSize = ceil(randomOutcomeLimit / randomDriverLimit);

do {
int bucket = nrand(randomDriverLimit);
int remainder = nrand(bucketSize);
randomOutcome = (bucket * bucketSize) + remainder;
} while (randomOutcome >= randomOutcomeLimit);
}


We compute bucketSize in a similar manner as Case 1, except that we use “round-up” (i.e. using the “ceil” function in C++) instead of “round-down”. We also ensure a non-fraction bucketSize (as explained in Case 1), by putting randomOutcomeLimit as the numerator, and randomDriverLimit as the denominator. Think of bucketSize as: “The number of randomOutcome per randomDriver”. 10 randomOutcomes per randomRandomDriver makes sense. 0.5 randomOutcome per randomRandomDriver does not.

We use “round-up” in the computation of the bucketSize to ensure the entire range of randomDriver can “cover” the entire range of randomOutcome. I can illustrate this point easily via some diagrams as followings.

In addition, we use nrand recursively downstream. Unfortunately I am not able to explain this bit in words. I would however recommend to look at the diagrams in the following section – looking at these diagrams have helped in understanding how the remaining steps hang together. I hope you may be able to figure this out for yourself too! (use the diagrams!).

### Further Visualisation

To visualise randomDriver, randomDriverLimit, randomOutcome, randomOutcomeLimit, bucket, and bucketSize all in one go, I have put together some PowerPoint diagrams here which I believe will speak for themselves without the need of much further (wordy) explanations! I assume a fixed randomDriverLimit of 2, and vary the randomOutcomeLimit for illustrations.

Visualise: randomDriverLimit of 2, randomOutcomeLimit = 5.

Visualise: randomDriverLimit of 2, randomOutcomeLimit = 6.

Visualise: randomDriverLimit of 2, randomOutcomeLimit = 7.

Visualise: randomDriverLimit of 3, randomOutcomeLimit = 5.

Visualise: randomDriverLimit of 3, randomOutcomeLimit = 6.

Visualise: randomDriverLimit of 3, randomOutcomeLimit = 7.

## The Project

To wrap up the entire projects I shall document the C++ header and source files here. The nrand algorithm aims to generate random numbers within the range [0, randomOutcomeLimit) in an even probability distribution – i.e. each random outcome has the same chance of being picked.

### Source Files

#### main.cpp

#include <iostream>    // std::cout, std::cin, std::endl
#include <cstdlib>     // RAND_MAX, srand
#include <ctime>       // std::time
#include "nrand.h"     // nrand

using std::cout;
using std::cin;
using std::endl;
using std::srand;
using std::time;

int main()
{
// Uncomment this line to see new randomness every time the job is run.
srand(time(NULL));

// Generate random numbers within the range [0, n)
const int n = 999999999;  // try n = 10, and n = 3276700
const int numbersPerLine = 5;

// Display the limits
cout << "randomOutcomeLimit: " << n << endl;
cout << "randomDriverLimit: " << (RAND_MAX + 1) << endl;

// Display the randomOutcomes
for (int i = 0; i != 100; ++i) {
if (i == 0)
cout << nrand(n);
else {
cout << ", " << nrand(n);

// start a new line
if (i != 0 && (i+1) % numbersPerLine == 0)
cout << endl;
}
}
return 0;
}


#### nrand.cpp

#include <cmath>         // ceil
#include <cstdlib>       // rand, RAND_MAX, srand
#include <stdexcept>     // domain_error

using std::ceil;
using std::rand;
using std::domain_error;

// Asymmetric ranges:
//   randomOutcome: [0, randomOutcomeLimit)
//   randomDriver:  [0, randomDriverLimit)

int nrand(int randomOutcomeLimit) {

if (randomOutcomeLimit <= 0)
throw domain_error("Argument to nrand is out of range");

int randomOutcome;
int randomDriverLimit = RAND_MAX + 1;

if (randomOutcomeLimit <= randomDriverLimit) {
const int bucketSize = randomDriverLimit / randomOutcomeLimit;

do {
int bucket = rand() / bucketSize;
randomOutcome = bucket;
} while (randomOutcome >= randomOutcomeLimit);

} else {
const int bucketSize = ceil(randomOutcomeLimit / randomDriverLimit);

do {
int bucket = nrand(randomDriverLimit);
int remainder = nrand(bucketSize);
randomOutcome = (bucket * bucketSize) + remainder;
} while (randomOutcome >= randomOutcomeLimit);
}
return randomOutcome;
}


#### nrand.h

#ifndef GUARD_NRAND_H
#define GUARD_NRAND_H

// nrand.h

int nrand(int);

#endif // GUARD_NRAND_H


## Test Result

### Test Case 1: when randomOutcomeLimit > randomDriverLimit

Set randomOutcomeLimit (i.e. the “n” in the main program) = 10. (randomDriverLimit on my Windows Vista System is 32768. i.e. RAND_MAX = 32767).

randomOutcomeLimit: 10
randomDriverLimit: 32768
0, 7, 3, 4, 5
, 2, 3, 5, 7, 8
, 5, 1, 0, 9, 1
, 0, 1, 2, 5, 3
, 1, 3, 5, 9, 1
, 8, 4, 8, 8, 8
, 2, 6, 8, 1, 4
, 5, 4, 5, 7, 4
, 7, 5, 1, 2, 0
, 5, 7, 3, 6, 9
, 8, 6, 5, 6, 4
, 4, 3, 2, 2, 5
, 9, 3, 0, 2, 7
, 8, 8, 1, 7, 1
, 4, 2, 3, 0, 0
, 8, 1, 2, 5, 9
, 4, 2, 5, 9, 9
, 7, 4, 3, 0, 4
, 9, 5, 6, 8, 0
, 4, 1, 6, 5, 2

Process returned 0 (0x0)   execution time : 0.293 s
Press any key to continue.


### Test Case 2: when randomOutcomeLimit <= randomDriverLimit

Set randomOutcomeLimit (i.e. the “n” in the main program) = 999999999. (randomDriverLimit on my Windows Vista System is 32768. i.e. RAND_MAX = 32767).

randomOutcomeLimit: 999999999
randomDriverLimit: 32768
40716497, 545497019, 401935189, 573265717, 87747530
, 675712555, 451019252, 549144813, 442862664, 914580320
, 3346215, 284749221, 212065917, 568356864, 979714558
, 544197549, 418473188, 720009720, 433822531, 424721581
, 541414236, 257758341, 608087599, 83473957, 984179693
, 492106188, 331214530, 521536020, 935375395, 580651497
, 195237728, 664468742, 414572965, 645131011, 682450472
, 477285531, 969901982, 319040147, 723433358, 897986272
, 120072600, 13688281, 426975024, 332152250, 774858930
, 85086071, 749114164, 49610236, 923599508, 602891952
, 599691044, 621946048, 156375628, 885810442, 918023686
, 135602107, 449128117, 661074165, 86314758, 699792585
, 461395423, 916830653, 398552621, 933505532, 326772882
, 142535635, 570358869, 573805944, 391791222, 527897950
, 942405333, 383272295, 777643889, 483575430, 421364976
, 808711544, 623226506, 83231569, 970354625, 918738412
, 272996097, 134992162, 663575905, 719215842, 172551438
, 345021838, 170806050, 154762077, 794127443, 949285447
, 81891829, 62877966, 418973269, 343240040, 888728945
, 627644733, 808847443, 814885588, 628243461, 188955197

Process returned 0 (0x0) execution time : 0.415 s
Press any key to continue.


# Reference

Koenig, Andrew & Moo, Barbara E., Accelerated C++, Addison-Wesley, 2000

# Exercise 7-8

Change the cross-reference program to find all the URLs in a file, and write all the lines on which each distinct URL occurs.

# Solution

This is surprisingly easy if you have completed the previous exercises. Chapter 7 of the book has provided the instruction to solve this. This exerise is about “doing it”.

This is the solution strategy:

1. Use the entire project as per Solution to Exercise 7-7. i.e. main.cpp, split.cpp/.h, xref.cpp/.h.
2. Add the find_url function components as per Solution to Exercise 6-0 (Part 4 / 7). i.e. find_urls.cpp/.h, not_url_char.cpp/.h, url_beg.cpp/.h, url_end.cpp/.h.
3. Update the main function so that this bit of code…
// Call xref using split by default.
map<string, vector<int> > ret = xref(cin);


… is replaced by this (using find_urls)….

// Call xref using find_urls (instead of the default split)
map<string, vector<int> > ret = xref(cin, find_urls);


Though split.cpp and split.h will not be called directly, the function split may is included by the xref function. So for safety, I will leave it in the project.

## The Project

In the following section I, for clarity sake, I shall include the entire project (C++ header and source files).

### Source Files

#### main.cpp

#include <iostream> // std::cin, std::cout, std::endl
#include <map> // std::map
#include <string> // std::string
#include <vector> // std::vector
#include "xref.h" // xref
#include "find_urls.h" // find_urls

using std::cin;
using std::cout;
using std::endl;
using std::map;
using std::string;
using std::vector;

// Find all the lines that refer to each word in the input
// (S7.3/128)

int main() {

// Call xref using find_urls (instead of the default split)
map<string, vector<int> > ret = xref(cin, find_urls);

// Write the results.
for (map<string, vector<int> >::const_iterator it = ret.begin(); it != ret.end(); ++it) {

// Find number of lines the word has appeared
vector<int>::size_type numLines = (it->second).size();

// Write the word
cout << it->first << " occurs on ";
if (numLines == 1)
cout << "line: ";
else
cout << "lines: ";

// Followed by one or more line numbers.
vector<int>::const_iterator line_it = it->second.begin();
cout << *line_it;   // write the first line number

++line_it;

// Write the rest of the line numbers, if any.
while (line_it != it->second.end()) {
cout << ", " << *line_it;
++line_it;
}

// Write a new line to separate each word from the next.
cout << endl;

}

return 0;
}



#### find_urls.cpp

#include <string> // string
#include <vector> // vector
#include "url_beg.h" // url_beg
#include "url_end.h" // url_end

using std::string;
using std::vector;

vector<string> find_urls(const string& s) {

vector<string> ret;
typedef string::const_iterator iter;
iter b = s.begin(), e = s.end();

// look through the entire input
while (b != e) {

// look for one or more letters followed by ://
b = url_beg(b, e);

// if we found it
if (b != e) {
// get the rest of the URL
iter after = url_end(b, e);

// remember the URL
ret.push_back(string(b, after));

// advance b and check for more URLs on this line
b = after;
}
} return ret;
}



#### not_url_char.cpp

#include <string> // string, isalnum
#include <algorithm> // find

using std::string;

bool not_url_char(char c) {

// characters, in addition to alphanumerics, that can appear in a URL
static const string url_ch = "~;/?:@=&\$-_.+!*'(),";

// see whether c can appear in a URL and return the negative
return !(isalnum(c) || find(url_ch.begin(), url_ch.end(), c) != url_ch.end() );

}



#### split.cpp

#include <string> // string
#include <vector> // vector
#include <cctype> // isspace
#include <algorithm> // find_if

using std::vector;
using std::string;
using std::isspace;
using std::find_if;

// true if the argument is whitespace, false otherwise
// (S6.1.1/103)
bool space(char c) { return isspace(c); }

// false if the argument is whitespace, true otherwise
// (S6.1.1/103)
bool not_space(char c) { return !isspace(c); }

// Scan a line and split into words. Return a vector that contains these words.
// (S6.1.1/103)
vector<string> split(const string& str) {
typedef string::const_iterator iter; vector<string> ret;

iter i = str.begin(); while (i != str.end()) {

i = find_if(i, str.end(), not_space);

// Find end of next word
iter j = find_if(i, str.end(), space);

// Copy the characters in ([i, j)
if (i != str.end())
ret.push_back(string(i, j));

// Re-initialize
i = j;

}

return ret;

}



#### url_beg.cpp

#include <string> // string, isalpha
#include <algorithm> // search
#include "not_url_char.h" // not_url_char

using std::string;

string::const_iterator url_beg(string::const_iterator b, string::const_iterator e) {

static const string sep = "://";
typedef string::const_iterator iter;

// i marks where the separator was found iter i = b;

while ((i = search(i, e, sep.begin(), sep.end() )) != e) {

// make sure the separator isn't at the beginning or end of the line
if (i != b && i + sep.size() != e) {

// beg marks the beginning of the protocol-name
iter beg = i;
while (beg != b && isalpha(beg[-1]))
--beg;

// is there at least one appropriate character before and after the separator?
if (beg != i && !not_url_char(i[sep.size()]))
return beg;
}

// the separator we found wasn't part of a URL; advance i past this separator
i += sep.size();

}

return e;

}


#### url_end.cpp

#include <string> // string
#include <vector> // vector
#include <algorithm> // find_if
#include "not_url_char.h" // not_url_char

using std::string;

string::const_iterator url_end(string::const_iterator b, string::const_iterator e) {
return find_if(b, e, not_url_char);
}



#### xref.cpp

#include <iostream> // std::cin, std::istream
#include <map> // std::map
#include <string> // std::string
#include <vector> // std::vector
#include <algorithm> // std::find
#include "split.h" // split

using std::cin;
using std::map;
using std::string;
using std::vector;
using std::istream;
using std::find;

// Find all the lines that refer to each word in the input
// In this program, the first line is represented by line 1,
// second line is represented by line 2, etc.
// (S7.3/126)
// Adjusted as per Exercise 7-3: store only non-duplicated line numbers.

map<string, vector<int> > xref(istream& in, vector<string> find_words(const string&) = split) {

string line; int line_number = 0; map<string, vector<int> > ret; // map string to vector<int>

// Read the next line while (getline(in, line)) { ++line_number;

// Break the input line into words
vector<string> words = find_words(line);

// remember that each word occurs on the current line
for (vector<string>::const_iterator it = words.begin();
it != words.end(); ++it) {

// store only non-duplicated line numbers.
if ( find( ret[*it].begin(), ret[*it].end(), line_number) == ret[*it].end() )
ret[*it].push_back(line_number);
}

}

return ret;
}



#### find_urls.h

#ifndef GUARD_FIND_URLS_H
#define GUARD_FIND_URLS_H
#include <vector>
#include <string>

std::vector<std::string> find_urls(const std::string&);

#endif // GUARD_FIND_URLS_H



#### not_url_char.h

#ifndef GUARD_NOT_URL_CHAR_H
#define GUARD_NOT_URL_CHAR_H

bool not_url_char(char);

#endif // GUARD_NOT_URL_CHAR_H



#### split.h

#ifndef GUARD_SPLIT_H
#define GUARD_SPLIT_H

#include <vector>
#include <string>

std::vector<std::string> split(const std::string&);

#endif // GUARD_SPLIT_H



#### url_beg.h

#ifndef GUARD_URL_BEG_H

#define GUARD_URL_BEG_H

std::string::const_iterator url_beg(std::string::const_iterator, std::string::const_iterator);

#endif // GUARD_URL_BEG_H



#### url_end.h

#ifndef GUARD_URL_END_H

#define GUARD_URL_END_H

#include <string>

std::string::const_iterator url_end(std::string::const_iterator, std::string::const_iterator);

#endif // GUARD_URL_END_H



#### xref.h

#ifndef GUARD_XREF_H

#define GUARD_XREF_H

// xref.h

#include <iostream> // std::istream
#include <map> // std::map
#include <string> // std::string
#include <vector> // std::vector
#include "split.h" // split

std::map<std::string, std::vector<int> >
xref(std::istream&, std::vector<std::string> find_words(const std::string&) = split);

#endif // GUARD_XREF_H



## Test Result

Run the program and submit some random text confirms the program works as expected.

hello world http://google.co.uk and http://bbc.co.uk
also http://bbc.co.uk
and http://yahoo.com
and finally http://w3schools.com
^Z
http://bbc.co.uk occurs on lines: 1, 3
http://google.co.uk occurs on lines: 1, 2, 5
http://w3schools.com occurs on line: 6
http://yahoo.com occurs on lines: 2, 4


# Reference

Koenig, Andrew & Moo, Barbara E., Accelerated C++, Addison-Wesley, 2000

# Exercise 7-7

Change the driver for the cross-reference program so that it writes line if there is only one line and lines otherwise.

# Solution

The solution to this exercise is surprisingly simple. To clarify the question a bit – it essentially asks us to have some sort of if-else condition in place.

For instance…

• If the word “apple” occurs only on one line, we print “apple occurs on line: ” (followed by the line number).
• Otherwise, if the word “apple” occurs on multiple lines, we print “apple occurs on lines: ” (followed by the line numbers).

Recall that the Solution to Exercise 7-3 already provides the skeleton project that reports non-duplicated word occurrence line numbers. It therefore makes sense to use that project as the baseline project, and do the adjustment on top of it.

It turns out that there is only one small bit of code that needs changing, and it resides at the main.cpp file (the main function).

This is the original snippet within the main function:

    // Write the word
cout << it->first << " occurs on line(s): ";


We can change it easily to the following to achieve that if-else conditioning. We essentially resolves the number of non-duplicate line occurences for the word beforehand. We can then use it to construct the if-else logic and display either “line” or “lines” accordingly.

    // Find number of lines the word has appeared
vector<int>::size_type numLines = (it->second).size();

// Write the word
cout << it->first << " occurs on ";
if (numLines == 1)
cout << "line: ";
else
cout << "lines: ";


Surprisingly simple.

## The Project

Simply…

1. Re-use that Project as per Solution to Exercise 7-3.
2. Replace the main.cpp file as per followings.
#include <iostream>   // std::cin, std::cout, std::endl
#include <map>        // std::map
#include <string>     // std::string
#include <vector>     // std::vector
#include "xref.h"     // xref

using std::cin;
using std::cout;
using std::endl;
using std::map;
using std::string;
using std::vector;

// Find all the lines that refer to each word in the input
// (S7.3/128)
int main()
{
// Call xref using split by default.
map<string, vector<int> > ret = xref(cin);

// Write the results.
for (map<string, vector<int> >::const_iterator it = ret.begin();
it != ret.end(); ++it) {

// Find number of lines the word has appeared
vector<int>::size_type numLines = (it->second).size();

// Write the word
cout << it->first << " occurs on ";
if (numLines == 1)
cout << "line: ";
else
cout << "lines: ";

// Followed by one or more line numbers.
vector<int>::const_iterator line_it = it->second.begin();
cout << *line_it;   // write the first line number

++line_it;
// Write the rest of the line numbers, if any.
while (line_it != it->second.end()) {
cout << ", " << *line_it;
++line_it;
}
// Write a new line to separate each word from the next.
cout << endl;
}

return 0;
}


## Test Result

Running this new version main code gives us the result as expected.

apple orange banana
orange orange
apple banana
mango
^Z
apple occurs on lines: 1, 3
banana occurs on lines: 1, 3
mango occurs on line: 4
orange occurs on lines: 1, 2


# Reference

Koenig, Andrew & Moo, Barbara E., Accelerated C++, Addison-Wesley, 2000

# Exercise 7-6

Reimplement the gen_sentence program using two vectors: One will hold the fully unwound, generated sentence, and the other will hold the rules and will be used as a stack. Do not use any recursive calls.

# Solution

After spending hours scratching my head and still getting stuck on this question, I decided to turn to Google for help. Eventually I discovered this Github Solution Code (uploaded by the Author himself) to this exercise. For the sake of learning, I simply picked out the “bits and pieces” from the Author’s code (such as the core concepts and snippets) and integrated these to my own solution. So let me disclose this upfront – the solution that you are seeing in this page is a result of the great help and pointer from the Author himself. Nevertheless, I believe it is the right thing to do to at least try and understand how the core concept work. The mission of this post is to add some additional documentation to demonstrate that we actually understand the concepts and techniques presented by the Author.

In the followings, I shall:

1. Disclose the project (i.e. the C++ source and header files) that I use in my learning.
2. Test run the job to confirm that it works.
3. Demonstrate that we understand the new concept by manually draw out the step by step logic.

## The Project

First, open the baseline project (i.e. the C++ source and header files) as per Solution to Exercise 7-0 (Part 3 / 3). Once the baseline project is opened up in (say) Code::Block, replace the following files with the newer version.

#### gen_aux.h

#ifndef GUARD_GEN_AUX_H
#define GUARD_GEN_AUX_H

// gen_aux.h

#include <string>             // std::string
#include <vector>             // std::vector

void gen_aux(const Grammar&, const std::string&, std::vector<std::string>&,
std::vector<std::string>&);

#endif // GUARD_GEN_AUX_H


#### gen_aux.cpp

#include <string>            // std::string
#include <vector>            // std::vector
#include <stdexcept>         // logic_error

#include "read_grammar.h"    // Grammar, Rule, Rule_collection
#include "bracketed.h"       // bracketed
#include "nrand.h"           // nrand

using std::string;
using std::vector;
using std::logic_error;

// Look up the input Grammar, and expand
// (modified version of S7.4.3/132)
void gen_aux(const Grammar& g, const string& token,
vector<string>& sentence, vector<string>& tokens) {
if (!bracketed(token)) {
sentence.push_back(token);
} else {
// locate the rule that corresponds to token'
Grammar::const_iterator it = g.find(token);

if (it == g.end())
throw logic_error("empty rule");

// fetch the set of possible rules
const Rule_collection& c = it->second;

// from which we select one at random
const Rule& r = c[nrand(c.size())];

// push rule's tokens onto stack of tokens
// (in reverse order, because we're pushing and popping from the back)
for (Rule::const_reverse_iterator i = r.rbegin(); i != r.rend(); ++i)
tokens.push_back(*i);
}
}


#### gen_sentence.cpp

#include <vector>             // std::vector
#include <string>             // std::string
#include "gen_aux.h"          // gen_aux

using std::vector;
using std::string;

// Generate a sentence based on a Grammar object
// (modified version of S7.4.3/132)
vector<string> gen_sentence(const Grammar& g) {
vector<string> sentence;
vector<string> tokens;
tokens.push_back("<sentence>");

while (!tokens.empty()) {
string token = tokens.back();
tokens.pop_back();
gen_aux(g, token, sentence, tokens);
}

return sentence;
}


## Test Result

Note: the result might differ when you run the job due to the nrand(). Core concept remains true though.

<noun>          cat
<noun>          dog
<noun>          table
<noun-phrase>   <noun>
<verb>          jumps
<verb>          sits
<location>      on the stairs
<location>      under the sky
<location>      wherever it wants
<sentence>      the <noun-phrase> <verb> <location>
^Z
the large brown dog sits wherever it wants
the table sits under the sky
the cat jumps on the stairs
the brown cat jumps under the sky
the brown large brown cat sits wherever it wants


## Understanding the concept

Let’s use the first output sentence to illustrate our understanding of the core concept used by the gen_sentence and gen_aux functions.

the large brown dog sits wherever it wants


The billion dollar question, how did the program generate this line?

To demonstrate my understanding, I’ve manually put together a couple of simple Excel tables to walk through the logic.

Recall that this is the input grammar table:

The gen_sentence and gen_aux can give the following output values:

### Some Notes To This Example

For some extra clarity, I include here a brief explanation of what each column means.

#### tokens non-empty?

If non-empty, the while loop inside the gen_sentence step would keep expanding tokens and construct the string sentence. Otherwise, the while loop stop and we have our sentence! (Note that we start by expanding the top level “<sentence>”)

vector<string> sentence;
vector<string> tokens;
tokens.push_back("<sentence>");
while (!tokens.empty()) {
// some codes to expand <sentence> and construct the string sentence.
}
return sentence;


#### const string& token

This is the last element of tokens, implied by this statement within the gen_sentence step:

string token = tokens.back();


#### Is token bracketed?

If the token is bracketed, it implies the token corresponds to a Grammar category and need expanding. If it is not bracketed, we  simply append the text the the string sentence.

if (!bracketed(token)) {
// some codes to construct the string sentence.
} else {
// some codes to expand the token.
}


#### const Rule& r

If the token is bracketed (i.e. a Grammar category), the const Rule& r is essentially the Grammar rule. I highlight this in orange.

#### vector<string>& sentence

If the token is not bracketed, we simply append the text to the string sentence. This column shows what the constructed string sentence look like. I highlight the newly appended element in blue.

#### (Updated) vector<string>& tokens

This is the interesting part and is updated by a combination of the gen_sentence and gen_aux steps.

• The tokens.pop_back() statement remove the last element of tokens (hence the strike-through in the table).
• If token (i.e. the token column) is bracketed, the for loop within the gen_aux function appends the entire rule (highlighted in orange) in reverse order. This is a very clever technique to get the tokens expand and string sentence constructed in the expected order. (it’s easier to visualise this via the Excel table above instead of explaning in writing!).

## Summary

Walking through the logic as illustrated in the Excel table, the program generates and return the complete string sentence. The actual sentence may get constructed differently due to the nrand().

# Reference

Koenig, Andrew & Moo, Barbara E., Accelerated C++, Addison-Wesley, 2000
The Github code solution uploaded by the author (A.Koening).

# Exercise 7-5

Reimplement the grammar program using a list as the data structure in which we build the sentence.

# Solution

This exercise is somewhat similar to Exercise 5-2 / 5-3 / 5-4, in which our objective is to replace all the relevant vector with list.

Here is the solution strategy:

1. Make a copy of the entire project as per Solution to Exercise 7-0 (Part 3 / 3) – i.e. the textbook Grammar / Generate Sentence program.
2. Replace all the vector with list, in main.cpp, gen_sentence.cpp/.h, gen_aux.cpp/.h.

## The main.cpp function


#include <vector>

// Some codes....

using std::vector;

// Some codes....

vector<string> sentence = gen_sentence(grammar);

// Some codes....

vector<string>::const_iterator it = sentence.begin();



We simply need to replace all the vector with list.

Here we can see the return type of the function gen_sentence needs to be changed from vector<string> to list<string>.

## The gen_sentence function

We need to replace all the vector with list in the gen_sentence.cpp (source file) and gen_sentence.h (header file).

gen_sentence.cpp


#include <vector>

// Some codes....

using std::vector;

// Some codes....

list<string> gen_sentence(const Grammar& g)
{
list<string> ret;
gen_aux(g, "<sentence>", ret);
return ret;
}



gen_sentence.h


// Some codes....

#include <vector>

std::vector<std::string> gen_sentence(const Grammar&);

// Some codes....



From this we see that we may need to adjust the gen_aux function to cater for the vector-to-list change.

## The gen_aux function

Looking at the gen_aux.cpp (source file) and gen_aux.h (header file) reveals that we need to replace the vector with list.

gen_aux.cpp


#include <vector>

// Some codes....

using std::vector;

// Some codes....

void gen_aux(const Grammar& g, const string& word, vector<string>& ret)
{
// Some codes....
}



gen_aux.h


// Some codes....

#include <vector>

void gen_aux(const Grammar&, const std::string&, std::vector<std::string>&);

// Some codes....



## The Project

Just simply make a copy of the entire project as per Solution to Exercise 7-0 (Part 3 / 3), and do the vector-to-list replacement as mentioned above.

i.e.  use the following (list version) files as replacement of the original  (vector version) files:

### Source File List (to change)

• main.cpp
• gen_sentence.cpp
• gen_aux.cpp

### Header File List (to change)

• gen_sentence.h
• gen_aux.h

### Source Files (to change)

#### main.cpp


#include <iostream>          // std::cin, std::cout, std::endl
#include <string>            // std::string
#include <list>              // std::list

#include "gen_sentence.h"    // gen_sentence

using std::cin;
using std::cout;
using std::endl;
using std::string;
using std::list;

using std::srand;
using std::time;

int main()
{
// Uncomment thhe below srand() to seed new randomness in new job-runs.
// srand(time(NULL));

for (int i = 0; i != 5; ++i) {

// generate the sentence
list<string> sentence = gen_sentence(grammar);

// write the first word, if any
list<string>::const_iterator it = sentence.begin();
if (!sentence.empty()) {
cout << *it;
++it;
}

// write the rest of the words, each preceded by a space
while (it != sentence.end()) {
cout << " " << *it;
++it;
}

cout << endl;
}

return 0;
}



#### gen_sentence.cpp


#include <list>             // std::list
#include <string>             // std::string
#include "gen_aux.h"          // gen_aux

using std::list;
using std::string;

// Generate a sentence based on a Grammer object
// (S7.4.3/132)
list<string> gen_sentence(const Grammar& g)
{
list<string> ret;
gen_aux(g, "<sentence>", ret);
return ret;
}



#### gen_aux.cpp


#include <string>            // std::string
#include <list>              // std::list
#include <stdexcept>         // logic_error

#include "read_grammar.h"    // Grammar, Rule, Rule_collection
#include "bracketed.h"       // bracketed
#include "nrand.h"           // nrand

using std::string;
using std::list;
using std::logic_error;

// Look up the input Grammar, and expand
// (S7.4.3/132)
void gen_aux(const Grammar& g, const string& word, list<string>& ret)
{
if (!bracketed(word)) {
ret.push_back(word);
} else {
// locate the rule that corresponds to word
Grammar::const_iterator it = g.find(word);
if (it == g.end())
throw logic_error("empty rule");

// fetch the set of possible rules
const Rule_collection& c = it->second;

// from which we select one at random
const Rule& r = c[nrand(c.size())];

// recursively expand the selected rule
for (Rule::const_iterator i = r.begin(); i != r.end(); ++i)
gen_aux(g, *i, ret);
}
}



#### gen_sentence.h


#ifndef GUARD_GEN_SENTENCE_H
#define GUARD_GEN_SENTENCE_H

// gen_sentence.h

#include <string>            // std::string
#include <list>              // std::list

std::list<std::string> gen_sentence(const Grammar&);

#endif // GUARD_GEN_SENTENCE_H



#### gen_aux.h


#ifndef GUARD_GEN_AUX_H
#define GUARD_GEN_AUX_H

// gen_aux.h

#include <string>             // std::string
#include <list>               // std::list

void gen_aux(const Grammar&, const std::string&, std::list<std::string>&);

#endif // GUARD_GEN_AUX_H



## Test Program

Following replace all the vector-to-list replacements, and resubmit the program with the same input, we get the results as expected. In my case:

<noun>          cat
<noun>          dog
<noun>          table
<noun-phrase>   <noun>
<verb>          jumps
<verb>          sits
<location>      on the stairs
<location>      under the sky
<location>      wherever it wants
<sentence>      the <noun-phrase> <verb> <location>
^Z
the large brown dog sits wherever it wants
the table sits under the sky
the cat jumps on the stairs
the brown cat jumps under the sky
the brown large brown cat sits wherever it wants


# Reference

Koenig, Andrew & Moo, Barbara E., Accelerated C++, Addison-Wesley, 2000

# Exercise 7-4

The output produced by the cross-reference program will be ungainly if the input file is large. Rewrite the program to break up the output if the lines get too long.

# Background

The original cross-reference program as per Solution to Exercise 7-0 (Part 2 / 3) lacks control in terms of output line length. For example, if my input “article” contains tons of repeated words, such as this:

apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple
orange orange orange orange orange orange orange orange orange orange orange orange orange orange orange orange
orange orange orange orange orange orange orange orange orange orange orange orange orange orange orange orange orange orange orange orange orange orange orange orange orange orange orange orange orange orange
orange orange orange orange orange orange orange
banana banana banana
banana
banana banana banana
banana banana
banana banana banana banana banana banana banana
banana banana
apple apple apple apple apple apple apple apple apple apple apple apple apple apple
apple apple apple apple apple apple apple apple apple apple apple apple apple apple
apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple apple


Given the above input, the default program produces output summary like this:

apple occurs on line(s): 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1
, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 13, 13, 13, 1
3, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13
banana occurs on line(s): 5, 5, 5, 6, 7, 7, 7, 8, 8, 9, 9, 9, 9, 9, 9, 9, 10, 10
orange occurs on line(s): 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3,
3, 4, 4, 4, 4, 4, 4, 4


The output lines may become very long – i.e. user may need to scroll left and right to see all the line numbers (not very user friendly). A bit “ungainly” as suggested by the author.

This therefore motivates us to add a control in place that ensures the output line length never exceeds a pre-defined value. i.e. if we set max line length to be 30 characters, and the output result contains 100 characters, the program should be able to split into 4 lines (first 3 lines contain 30 characters, and the 4th line contains 10 characters). This way, the user won’t need to scroll left and right to see all line numbers – more user friendly.

The following section will describe a solution to this problem.

# Solution

This is our solution strategy:

1. Make a copy of the project (i.e. source and header files) as per Solution to Exercise 7-0 (Part 2 / 3).
2. Update the main program to incorporate the desirable line-length control.

## Incorporate Line-length Control

First of all, we need define a const string::size_type lineLength, which is the maximum line-length of the output.

We then need to have a means to concatenate various output values.

In the Solution to Exercise 5-1, we learnt that the variable type std::ostringstream (of <sstream>  directive) is way more superior in concatenation in comparison to std::string (of <string> directive). i.e. there seem to be much less constraints in terms of concatenating values of different types for a std::ostringstream than std::string. For this reason, we shall use std::ostringstream to store output. Note also that ostreamstring can be converted into a string easily.

When displaying the line numbers for each word, we implement a if-condition to ensure the line length doesn’t exceed our pre-defined lineLength. The modulus operator (%) will come in handy for this.

## The Project

Just simply re-use the entire project from Solution to Exercise 7-0 (Part 2 / 3), and replace the main program with the following. (Note that for illustration sake the pre-defined lineLength is set to 60 characters max per line).


#include <iostream>   // std::cin, std::cout, std::endl
#include <map>        // std::map
#include <string>     // std::string
#include <vector>     // std::vector
#include <sstream>    // std::ostringstream
#include "xref.h"     // xref

using std::cin;
using std::cout;
using std::endl;
using std::map;
using std::string;
using std::vector;
using std::ostringstream;

// Find all the lines that refer to each word in the input
// (S7.3/128)
int main()
{
// Call xref using split by default.
map<string, vector<int> > ret = xref(cin);

// Set the width of output line to this max width limit.
const string::size_type lineLength = 60;

// Write the results.
for (map<string, vector<int> >::const_iterator it = ret.begin();
it != ret.end(); ++it) {

// We use ostreamstring for its powerful concatenation properties.
// We can pretty much concatenate anything to an ostreamstring, in
// comparison to string.
ostringstream outputStream;

// Write the word
outputStream << it->first << " occurs on line(s): ";

// Followed by one or more line numbers.
vector<int>::const_iterator line_it = it->second.begin();
outputStream << *line_it;
++line_it;

// Write the rest of the line numbers, if any.
while (line_it != it->second.end()) {
outputStream << ", " << *line_it;
++line_it;
}

// Break outputStream into multiple lines with max width of lineLength.
string outputLine = outputStream.str();
for (string::size_type i = 0; i != outputLine.size(); ++i ) {
if (i % lineLength == 0) {
cout << endl;
}
cout << outputLine[i];
}

// Write a new line to separate each word from the next.
cout << endl;
}

return 0;
}



## Test Program

Re-submitting the same input (see the top of this post), our output now looks much more “controlled”, in terms of line length.


apple occurs on line(s): 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1,
1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 1, 11, 1
1, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 11, 12, 12, 1
2, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 12, 13, 13, 13, 1
3, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 13, 1
3, 13, 13, 13, 13, 13, 13, 13, 13, 13

banana occurs on line(s): 5, 5, 5, 6, 7, 7, 7, 8, 8, 9, 9, 9
, 9, 9, 9, 9, 10, 10

orange occurs on line(s): 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2, 2
, 2, 2, 2, 2, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3
, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 3, 4, 4, 4, 4, 4, 4
, 4



Not perfect, but better.

# Reference

Koenig, Andrew & Moo, Barbara E., Accelerated C++, Addison-Wesley, 2000

# Exercise 7-3

The cross-reference program from S7.3/126 could be improved: As it stands, if a word occurs more than once on the same input line, the porgram will report that line multiple times. Change the code so that it detects multiple occurences of the same line number and inserts the line number only once.

## Background

Remember that when we run the Chapter 7 cross-reference program (i.e. Solution to Exercise 7-0 (Part 2 / 3) ), we have the following test results:

apple orange banana
orange orange
apple banana
mango
^Z
apple occurs on line(s): 1, 3
banana occurs on line(s): 1, 3
mango occurs on line(s): 4
orange occurs on line(s): 1, 2, 2


## Note:

1. Recall that the xref program assumes the first line as line 1 (instead of 0), second line as line 2 (instead of 1). This is as designed.
2. The word “orange” appears on line 2 twice. So when it comes to being displayed in the summary, it shows the line number “2” twice.
3. The objective of this exercise is to only store non-duplicated line number.

## Solution Strategy

We only need to make a very small enhancement – to adjust the xref function by adding a small bit of code. Include a if condition to check whether the line number already exists for the corresponding associative array. Use the find utility of the <algorithm> directive to help us do this.

### The Project

We can reuse the entire project as per Solution to Exercise 7-0 (Part 2 / 3). Simply replace the xref.cpp file with this new one:

#### xref.cpp


#include <iostream>   // std::cin, std::istream
#include <map>        // std::map
#include <string>     // std::string
#include <vector>     // std::vector
#include <algorithm>  // std::find
#include "split.h"    // split

using std::cin;
using std::map;
using std::string;
using std::vector;
using std::istream;
using std::find;

// Find all the lines that refer to each word in the input
// In this program, the first line is represented by line 1,
// second line is represented by line 2, etc.
// (S7.3/126)
// Adjusted as per Exercise 7-3: store only non-duplicated line numbers.
map<string, vector<int> >
xref(istream& in, vector<string> find_words(const string&) = split)
{
string line;
int line_number = 0;
map<string, vector<int> > ret;  // map string to vector<int>

while (getline(in, line)) {
++line_number;

// Break the input line into words
vector<string> words = find_words(line);

// remember that each word occurs on the current line
for (vector<string>::const_iterator it = words.begin();
it != words.end(); ++it) {

// store only non-duplicated line numbers.
if ( find( ret[*it].begin(), ret[*it].end(), line_number) == ret[*it].end() )
ret[*it].push_back(line_number);
}
}

return ret;
}



# Test Program

Re-submitting the program with the same input, the program now produces non-duplicated line numbers per associative array (word).

apple orange banana
orange orange
apple banana
mango
^Z
apple occurs on line(s): 1, 3
banana occurs on line(s): 1, 3
mango occurs on line(s): 4
orange occurs on line(s): 1, 2


# Reference

Koenig, Andrew & Moo, Barbara E., Accelerated C++, Addison-Wesley, 2000

# Exercise 7-2

Extend the program in S4.2.3/64 to assign letter grades by ranges:

A 90-100
B 80-89.99...
C 70-79.99...
D 60-69.99...
F < 60


# Solution

The strategy:

1. Make a copy of the entire projects from Solution to Exercise 4-0. (i.e. the sample programs in Chapter 4 of the textbook).
3. In the main program, use the getletterGrade function to compute the student’s string letterGrade. Have an associative array letterGradeCount that maps a string (letter grade) to integer (to keep track of letter grade occurrences).

## The Program

Here summarises the entire Code::Block Project, including the header and source files.

### Source Files

#### main.cpp


#include <algorithm>
#include <iomanip>
#include <ios>
#include <iostream>
#include <stdexcept>
#include <string>
#include <vector>
#include <map>
#include "Student_info.h"

using std::cin;
using std::cout;
using std::endl;
using std::domain_error;
using std::max;
using std::setprecision;
using std::sort;
using std::streamsize;
using std::string;
using std::vector;
using std::map;

int main()
{
vector<Student_info> students;
Student_info record;
string::size_type maxlen = 0;   // the length of the longest name

// read and store all the student's data.
// Invariant:   students contain all the student records read so far
//              maxlen contains the length of the longest name in students
{
// find the length of longest name
maxlen = max(maxlen, record.name.size());
students.push_back(record);
}

// alphabetize the student records
sort(students.begin(), students.end(), compare);

// write the names and grades
for (vector<Student_info>::size_type i = 0;
i != students.size(); ++i)
{
//write the name, padded on teh right to maxlen + 1 characters
cout << students[i].name
<< string(maxlen + 1 - students[i].name.size(), ' ');

try
{
streamsize prec = cout.precision();
<< setprecision(prec) << " (" << letterGrade << ")";
}
catch (domain_error e)
{
cout << e.what();
}
cout << endl;
}
cout << "\nLetter Grade Summary" << endl;
for (map<string, int>::const_iterator i = letterGradeCount.begin();
cout << i->first << ": " << i->second << endl;

return 0;
}




#include <string>

using std::string;

// (For Exercise 7-2)

string ret;

ret = "A";
ret = "B";
ret = "C";
ret = "D";
else
ret = "F";

return ret;
}




#include <stdexcept>
#include <vector>
#include "median.h"
#include "Student_info.h"

using std::domain_error;
using std::vector;

// definitions for the grade functions from S4.1/52, S4.1.2/54, S4.2.2/63

// compute a student's overall grade from midterm and final exam
double grade(double midterm, double final, double homework)
{
return 0.2 * midterm + 0.4 * final + 0.4 * homework;
}

// compute a student's overall grade from midterm and final exam grades
// and vector of homework grades.
// this function does not copy its argument, because median (function) does it for us.
// (S4.1.2/54)
double grade(double midterm, double final, const vector<double>& hw)
{
if (hw.size() == 0)
throw domain_error("student has done no homework");
}

// this function computes the final grade for a Student_info object
// (S4.2.2/63)
{
}


#### median.cpp

// source file for the median function
#include <algorithm>
#include <stdexcept>
#include <vector>

using std::domain_error;
using std::sort;
using std::vector;

// compute the median of a vector<double>
// (S4.1.1/53)
double median(vector<double> vec)
{
typedef vector<double>::size_type vec_sz;

vec_sz size = vec.size();
if (size == 0)
throw domain_error("median of an empty vector");

sort(vec.begin(),vec.end());

vec_sz mid = size/2;

return size % 2 == 0 ? (vec[mid] + vec[mid-1]) / 2 : vec[mid];
}



#### Student_info.cpp


#ifndef GUARD_STUDENT_INFO_H
#define GUARD_STUDENT_INFO_H

// Student_info.h
#include <iostream>
#include <string>
#include <vector>

struct Student_info
{
std::string name;
double midterm, final;
std::vector<double> homework;
};

bool compare(const Student_info&, const Student_info&);

#endif // GUARD_STUDENT_INFO_H

</p>

#ifndef GUARD_GETLETTERGRADE_H

#include <string>





#include <vector>
#include "Student_info.h"



#### median.h

#ifndef GUARD_MEDIAN_H
#define GUARD_MEDIAN_H

// median.h - final version
#include <vector>
double median(std::vector<double>);

#endif // GUARD_MEDIAN_H



#### Student_info.h


#ifndef GUARD_STUDENT_INFO_H
#define GUARD_STUDENT_INFO_H

// Student_info.h
#include <iostream>
#include <string>
#include <vector>

struct Student_info
{
std::string name;
double midterm, final;
std::vector<double> homework;
};

bool compare(const Student_info&, const Student_info&);

#endif // GUARD_STUDENT_INFO_H



## Test Program

I test the program by submitting a list of sample student grades. The program automatically compute and display the numeric and letter grades. It also creates a letter grade summary to count occurances per letter grade.

johnny   90 90 90 90 90
jason    85 85 85 85 85
billy    82 82 82 82 82
fred     80 80 80 80 80
frank    70 70 70 70 70
jay      66 66 66 66 66
bob      60 60 60 60 60
mary     50 50 50 50 50
ann      40 40 40 40 40
bill     30 30 30 30 30
goat     10 10 10 10 10
dog      0  0  0  0  0
^Z
^Z
ann    40 (F)
bill   30 (F)
billy  82 (B)
bob    60 (D)
dog    0 (F)
frank  70 (C)
fred   80 (B)
goat   10 (F)
jason  85 (B)
jay    66 (D)
johnny 90 (A)
mary   50 (F)

`